1. **State the problem:** We need to find all values of $x$ such that the triangle with vertices $A(x,4,-1)$, $B(0,4,-3)$, and $C(2,1,1)$ has a right angle at vertex $A$. 2. **Recall the condition for a right angle at $A$:** The vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ must be perpendicular. This means their dot product is zero: $$\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$$ 3. **Find vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:** $$\overrightarrow{AB} = B - A = (0 - x, 4 - 4, -3 - (-1)) = (-x, 0, -2)$$ $$\overrightarrow{AC} = C - A = (2 - x, 1 - 4, 1 - (-1)) = (2 - x, -3, 2)$$ 4. **Compute the dot product:** $$\overrightarrow{AB} \cdot \overrightarrow{AC} = (-x)(2 - x) + 0 \cdot (-3) + (-2)(2) = -x(2 - x) - 4$$ 5. **Set the dot product equal to zero and solve for $x$:** $$-x(2 - x) - 4 = 0$$ $$-2x + x^2 - 4 = 0$$ $$x^2 - 2x - 4 = 0$$ 6. **Use the quadratic formula:** $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2} = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2}$$ 7. **Simplify the square root:** $$\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$$ 8. **Final solutions:** $$x = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}$$ **Answer:** $$x = 1 + \sqrt{5}, 1 - \sqrt{5}$$